I need to prove the following, but I am not able to do it. This is not homework, nor something related to research, but rather something that came up in preparation for an exam.

If $n = 1 + m$, where $m$ is the product of four consecutive positive
integers, prove that $n$ is a perfect square.

Now since $m = p(p+1)(p+2)(p+3)$;

$p = 0, n = 1$ - Perfect Square

$p = 1, n = 25$ - Perfect Square

$p = 2, n = 121$ - Perfect Square

Is there any way to prove the above without induction? My approach was to expand $m = p(p+1)(p+2)(p+3)$ into a 4th degree equation, and then try proving that $n = m + 1$ is a perfect square, but I wasn't able to do it. Any idea if it is possible?

11 Answers
11

Your technique should have worked, but if you don't know which expansions to do first you can get yourself in a tangle of algebra and make silly mistakes that bring the whole thing crashing down.

The way I reasoned was, well, I have four numbers multiplied together, and I want it to be two numbers of the same size multiplied together. So I'll try multiplying the big one with the small one, and the two middle ones.

$$p(p+1)(p+2)(p+3) + 1 = (p^2 + 3p)(p^2 + 3p + 2) + 1$$

Now those terms are nearly the same. How can we force them together? I'm going to use the basic but sometimes-overlooked fact that $xy = (x+1)y - y$, and likewise $x(y + 1) = xy + x$.

Wow you did it, but tell me one thing.You said "I have four numbers multiplied together, and I want it to be two numbers of the same size multiplied together. ", but why? I mean 1 is still being added to expression.
–
Kartik AnandJun 7 '12 at 9:50

7

Well, I was kind of ignoring the 1 for the time being; I wanted two numbers roughly the same, and I thought the details likely to work themselves out :)
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Ben MillwoodJun 7 '12 at 9:56

7

Slightly smoother: When you are at $(p^2+3p)(p^2+3p+2)+1$, let $x=p^2+p+1$, We are looking at $(x-1)(x+1)+1=x^2-1+1=x^2$.
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André NicolasJun 7 '12 at 10:14

2

@AndréNicolas make it 3p in the expression for x
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Kartik AnandJun 7 '12 at 10:19

9

This goes by what I like to call the 'Mathematics of wishful thinking' : there is a solution, we sort of know what it should look like, and so we go and hope that everything works out. +1
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mixedmath♦Jun 7 '12 at 19:29

I like this method. The first thing you do is try to make the equation at the top more symmetrical, which was basically my idea, but we went about it in different ways.
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Ben MillwoodJun 7 '12 at 10:10

@benmachine totally agree, but I just don't that like the number 1.5
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zinkingJun 8 '12 at 2:18

You miss one final step; to see that it is a perfect square of an integer, you need that $x^2 - 5/4$ is an integer. In this case you could say you are "lucky" that $2 \cdot (3/2), (3/2)^2 - 5/4 \in \mathbb{N}$.
–
TMMJun 7 '12 at 19:31

1

@TMM: Or you could just say that it's an integer which is the perfect square of a rational number, which makes it the perfect square of an integer.
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MicahJun 7 '12 at 23:05

Yes, I did miss the final step, but there's no luck involved - it's straightforward algebra that x^2 - 5/4 is an integer. See my edit which completes the proof.
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JTBJun 8 '12 at 13:02

I think there are two issues here. One is constructing the quartic, which just depends on you doing the algebra correctly. The second is proceeding to factorise the quartic. It would be easier to factorise it if you know what the factorisation is going to be.

To discover this, I tried a few examples. For $p=7$, the quartic gives $5041=71^2$. For $p=14$, the quartic gives $57121=239^2$. I noticed that $71=72-1=8\times9-1$ and $239=15\times16-1$.

Note that $p^4$ is $(p^2)^2$, so this is equal to a square plus something extra. If this is to be a square number, then the extra must be a sum of odd numbers starting with $2p^2+1$, that is, we must have

$$6p^3+11p^2+6p+1=\sum^n_{k=0}(2(p^2+k)+1)$$

for some $n$. That sum can be computed to be

$$(n+1)(2p^2+1)+(n+1)n$$

and now it's quite easy to see that there's only one possible choice for $n$. Indeed, we want this to be a cubic in $p$, so $n$ must be linear in $p$. The coefficient of the linear term must be $3$, so that we get a cubic term of $6p^3$ after multiplying out. And since we want a constant term of $1$, we see that there can be no constant term in $n$. So $n=3p$ is the only possible choice. Plugging this in, we find that it works.